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anonymous
 5 years ago
Is anyone good with vector space?
anonymous
 5 years ago
Is anyone good with vector space?

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nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0They are my homies :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ha, ha, ha. So are u down with them as well?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Really though tell them to ease up a little. They are beating up my brain......

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Any concrete question?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah. I am having a hard time understanding this concept of going down the list of axioms to prove whether or not they make the statement true.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here is an example: determine whether the given set of vectors is closed under addition and closed under scalar multiplication. The set of scalars is the set of all real numbers. The set of all solutions to the diff eqn. y'+9y=ax^2

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Well, there are a lot of nice properties of and theorems relating to vector spaces. To know that they apply you have to know whether some set really is a vector space. And for that all those axioms have to be true. Basically it's a kind of linear structure.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, thats sounds similar to what my teacher says that you have to show that they are all true. I am having trouble proving the axioms. I guess I just have to look at examples and figure it out.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0For solutions of differential equations you normally need a homogeneous form. In your example that is not the case and you can see that the solutions don't form a vector space.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For the problem, the ax^2 is really 4x^2. sorry for the typo.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0Still the same. Just asume you have a solution f: x ↦ y and look whether for any c the solution c*f is still a solution. You will see, it's not.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if it were to equal zero than it would be one?

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.0yes, because the differentiation is linear as you from the basic differentiation rules.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0O.k thanks. Your a life saver.
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